Chemistry 111 Exam 4 Chemical Reactions in Solution Precipitation Reactions o Because water is polar it has the ability to dissolve a large number of substances Solutions that have water as a solvent are called aqueous aq o When an ionic compound dissolves in water it dissociates into separate ions This allows the ions to move freely in water o Precipitation Reactions occur when two soluble ions meet to form an insoluble precipitate aka solid s o Precipitation reactions are often a type of double replacement reaction Chemical Reactions in Solution Ionic Compounds o When solutions of salts are mixed a solid forms if ions of an insoluble salt are present o Example o Ionic Compounds that AgNO3 aq NaCl aq AgCl s NaNO3 aq dissolve in water are soluble salts do not dissolve in water are insoluble salts Soluble salts typically contain at least one ion from Groups 1 A 1 NO3 or C2H3O2 acetate Most other combinations are insoluble Chemical Reactions in Solution Solubility Rules Chart provided on tests Learning Check o Use the solubility rules to determine if each salt is S soluble or I insoluble Explain Na2SO4 MgCO3 PbCl2 Soluble Insoluble Insoluble 1 P a g e Chemistry 111 Exam 4 Soluble MgCl2 o For each reaction in solution predict the products and determine their solubility BaCl2 aq Na2SO4 aq BaSO4 s 2NaCl aq AgNO3 aq KCl aq AgCl s KNO3 aq KNO3 aq NaCl aq KCl aq NaNO3 aq No reaction all soluble o Are the following soluble of insoluble Na2SO4 PbSO4 Li2CO3 MgCl2 Soluble Soluble Insoluble Soluble Concentration Units Weight Volume Percent Concentration o The concentration of a solution tells how much solute is dissolved in a given amount of solution o Weight volume percent concentration w v is the number of grams of solute dissolved in 100 mL of solution o For example vinegar contains 5 g of acetic acid in 100 mL of w v mass of solute g volume of solution mL 100 solution so the w v concentration is 5 gacetic acid 100 mLVinegar solution 100 5 w v Concentration Units Using a percent concentration as a conversion factor o Sample problem A saline solution used in intravenous drips contains 0 92 w v NaCl in water How many grams of NaCl are contained in 250 mL of this solution Step 1 Identify the known quantities and the desired quantity Step 2 0 92g NaCl 100mLsolution 0 92 w v NaCl solution 250mL Known quantities g NaCl desired quantity Write out the conversion factors Step 3 Solve the problem 250 mL 0 92 g NaCl 100 mL solution 2 3 g NaCl Concentration Units Parts Per Million 2 P a g e Chemistry 111 Exam 4 o When a solution contains a very small concentration of solute it is often expressed in parts per million that s where the 106 comes from ppm 106 mass of solute g mass of solution g OR ppm mass of solute mL mass of solution mL 106 Chemistry Question What is hard water o I m sure you ve heard the terms hard water and soft water but do you know what they mean Hard Water is any water containing an appreciable quantity of dissolved minerals These ions enter the water when acidic ground water dissolves mineral deposits in the Earth s crust o Mineral deposits on your cooking dishes or rings of insoluble soap scum in your bathtub are signs of hard water The dissolved minerals in hard water contain cations with a charge of 2 usually Ca2 Mg2 and Fe2 Soap Scum soaps react with hardness ions to form insoluble compounds i e a precipitate or solid scum 2 CH3 CH2 16CO2 Na Mg2 CH3 CH2 16CO2 2Mg2 2 Na o Sodium carbonate is added to many detergents as a water softening agent o The doubly positive calcium and magnesium ions of hard water preferentially bind with the doubly negative carbonate ion freeing the detergent molecules to do their job Concentration Units Molarity o Molarity is the number of moles of solute per liter of solution abbreviated as M Molarity M molesof solute mol liter of solution L o Since the number of grams and moles of a substance is related by the molar mass we can convert a given volume of solution to the number of grams of solute given its molarity How to calculate molarity from a given number of grams of solute o Example Calculate the molarity of a solution made from 20 0 g of NaOH in 250 mL of solution Step 1 Identify the known quantities and the desired quantity 20 0 g NaOH 3 P a g e Chemistry 111 Exam 4 250 mL solution Known Quantities M mol L Desired Quantity Step 2 Use the molar mass to convert grams of NaOH to moles of NaOH molar mass 40 0 g mol 1mol 40 0 g NaOH 0 500 mol NaOH 20 0 g NaOH Convert milliliters of solution to liters of solution using a mL L conversion factor 250 ml solution 1 L 0 25 Lsolution 1000 mL Step 3 Divide the number of moles of solute by the number of liters of solution to obtain the molarity M moles of solute mol V L 0 500 mol NaOH 0 25 L solution 2 0M 2 0 mol L Concentration Units Molarity o Molarity is a conversion factor that relates the number of moles of solute to the volume of solution it occupies o To calculate the moles of solute rearrange the equation for molarity moles of solute mol M V L o To calculate the volume of the solution rearrange the equation for molarity V L molesof solute mol M o Sample Problem What volume in milliliters of a 0 30 M 0 30mol L solution of glucose contains 0 025 mol of glucose Step 1 Identify the known quantities and the desired quantity 0 30 M 0 025 mol glucose Known Quantities V L Desired Quantity Step 2 Divide the number of moles by molarity to obtain the volume in liters V L molesof solute mol M 0 025 mol glucose 0 30mol L Step 3 Use a mL L conversion factor to convert liters to milliliters 0 083 Lsolution 1000 mol 83mLglucose solution 1L Learning Check o How many milliliters of 2 00 M HNO3 contain 24 0 g of HNO3 4 P a g e Chemistry 111 Exam 4 24 0 g HNO3 1 mol HNO 3 63 0 g HNO3 HNO Molar Mass 63 0g mol 1000 mL 1L solution 2 00 mol HNO3 1 L 190 mLof 2 00 M HNO 3 Dilution o Dilution is the process of reducing the concentration of a solute in solution by adding more solvent o Dilution is the addition of solvent to decrease the concentration C of solute The solution volume changes but the amount of solute is constant moles of solute mol molarity M x volume V C1V1 initial values C2V2 final values o Example How many milliliters of a 4 0 w v solution must be used to prepare 250 mL of a 0 080 w v solution Identify …